MATH 308 Lecture 31

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Begin Exam 4 content

Eigenvalues and Eigenvectors

$A = [\begin{matrix} 2 & - 3 \\ 1 & - 2 \end{matrix}]$

$λ \in {- 1, 1}$

$λ = 1 : \vec{v_{1}} = ⟨ 3, 1 ⟩$

$λ = - 1 : \vec{v_{2}} = ⟨ 1, 1 ⟩$

Find the General solution of $X^{'} (t) = A X$ and describe the behavior as $t \to \infty$ .

We need to find two linearly independent solutions and take a linear combination of them.

Let $X_{2} (t) = f_{2} (t) V_{2}$ , where $f_{2}$ is any scalar function, and $V_{2}$ is the eigenvector for $λ = - 1$ .

So for our diff eq, ${X_{2}}^{'} (t) = A X_{2} (t)$

We get

$\begin{aligned} {f_{2}}^{'} (t) V_{2} & = f_{2} (t) (A V_{2} (t)) \\ = f_{2} (t) (- 1) V_{2} \end{aligned}$

Recall that $(A - (- 1) I) V_{2} = 0$ , so $A V_{2} = (- 1) V_{2}$ . That's where the second form came from.

If we plug in $V_{2} = ⟨ 1, 1 ⟩$ , we get

$[\begin{matrix} {f_{2}}^{'} (t) \\ {f_{2}}^{'} (t) \end{matrix}] = [\begin{matrix} - f_{2} (t) \\ - f_{2} (t) \end{matrix}]$

Thus ${f_{2}}^{'} (t) = - f_{2} (t)$ , and the solution to this differential equation is $f_{2} (t) = C e^{- t}$

Putting this back in Matrix form, we get

$X_{2} (t) = C e^{- t} [\begin{matrix} 1 \\ 1 \end{matrix}]$

A similar process for $V_{1}$ gives

$f_{1} (t) = C e^{t}$

And

$X_{1} (t) = C e^{t} [\begin{matrix} 3 \\ 1 \end{matrix}]$

Therefore, the general solution is

[\begin{matrix} x (t) \\ y (t) \end{matrix}] = c_{1} e^{t} [\begin{matrix} 3 \\ 1 \end{matrix}] + c_{2} e^{- t} [\begin{matrix} 1 \\ 1 \end{matrix}]